A rigid container holds a gas at a pressure of 55kPa and a temp of -100.0 deg C. What will the pressure be when the temp is increased to 200.0 deg C?

To solve this problem, we can use the Gay-Lussac's Law formula, which relates pressure and temperature of a gas in a rigid container:

P1/T1 = P2/T2

where P1 is initial pressure, T1 is initial temperature in Kelvin, P2 is final pressure, and T2 is final temperature in Kelvin.

First, let's convert the temperatures to Kelvin:
T1 = -100.0 + 273.15 = 173.15 K
T2 = 200.0 + 273.15 = 473.15 K

Now, plug the values into the equation and solve for P2:

55 kPa / 173.15 K = P2 / 473.15 K

P2 = (55 kPa * 473.15 K) / 173.15 K = 148.75 kPa

So, the final pressure will be approximately 148.75 kPa when the temperature is increased to 200.0°C.

To find out the pressure when the temperature is increased, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (in Pascal)
V = Volume (in m^3)
n = Number of moles
R = Ideal gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)

To solve this problem, we need to convert the given values to Kelvin and find the final pressure by using the ideal gas law equation.

Given:
Initial Pressure (P1) = 55 kPa
Initial Temperature (T1) = -100.0 °C = -100.0 + 273.15 = 173.15 K
Final Temperature (T2) = 200.0 °C = 200.0 + 273.15 = 473.15 K

Assuming the volume and the number of moles of gas remain constant, we can rearrange the ideal gas law equation to solve for the final pressure (P2):

P1/T1 = P2/T2

Substituting the values:

55 kPa / 173.15 K = P2 / 473.15 K

Now, let's solve for P2:

P2 = (55 kPa * 473.15 K) / 173.15 K
P2 = 150.41 kPa

Therefore, when the temperature is increased to 200.0 °C, the pressure will be approximately 150.41 kPa.

To determine the pressure of the gas when the temperature is increased to 200.0 degrees Celsius, we can use the ideal gas law. The ideal gas law is expressed as:

PV = nRT

Where:
P is the pressure of the gas,
V is the volume of the container,
n is the number of moles of the gas,
R is the ideal gas constant, and
T is the temperature of the gas in Kelvin.

First, we need to convert the initial temperature of -100.0 degrees Celsius to Kelvin:

T1 = -100.0 + 273.15
T1 = 173.15 K

The initial pressure, P1, is given as 55 kPa.

Next, we need to convert the final temperature of 200.0 degrees Celsius to Kelvin:

T2 = 200.0 + 273.15
T2 = 473.15 K

Now, we can rearrange the ideal gas law equation to solve for the final pressure, P2:

P1V1/T1 = P2V2/T2

Since the volume and the number of moles of gas are constant in this case, we can simplify the equation to:

P1/T1 = P2/T2

Now, plug in the values we have:

P1 = 55 kPa
T1 = 173.15 K
T2 = 473.15 K

P2/T2 = P1/T1

P2 = P1 * (T2 / T1)
P2 = 55 * (473.15 / 173.15)
P2 ≈ 150.65 kPa

Therefore, when the temperature is increased to 200.0 degrees Celsius, the pressure of the gas in the rigid container will be approximately 150.65 kPa.